Problem 19: Balkan MO Shortlist 2007

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by henderson, Jun 7, 2016, 6:04 PM

$\bf\color{red}Problem\ 19\$ Let $\rho(O)$ be a circle and $A$ a point outside it. Denote by $B,C$ the points where the tangents from $A$ with respect to $\rho(O)$ meet the circle, $D$ the point on $\rho(O),$ for which $O\in AD,$ $X$ the foot the perpendicular from $B$ to $CD,$ $Y$ the midpoint of the line segment $BX$ and by $Z$ the second intersection of $DY$ with $\rho(O).$ Prove that $ZA\perp ZC.$
$\bf\color{red}My\ solution\$ According to this problem, we have $\angle BZS+\angle CZA=180^\circ.$ So, $\angle CZA=90^\circ$ $\iff$ $\angle BZS=90^\circ.$ Also, $\angle BZS=90^\circ$ $\iff$ $BZSY$ is a cyclic quadrilateral (where $S$ is the midpoint of $BC$ ). But, since $\frac{BS}{BC}=\frac{BY}{BX},$ we get $\triangle BSY \sim \triangle BCX,$ so $\angle BSY=\angle BCX.$ Because $\angle BCX=\angle BZY$ (in cyclic quadrilateral $BZCD$ ) $,$ we conclude $\angle BSY=\angle BZY,$ which means that $BZSY$ is a cyclic quadrilateral. $\square$

This post has been edited 7 times. Last edited by henderson, Sep 25, 2016, 2:49 PM

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Problem 19: Balkan MO Shortlist 2007

by henderson, Jun 7, 2016, 6:04 PM

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